let's try a bit of useless [this side of Vegas at least] calculations.

In a standard deck of cards there are 52 cards: 1-10, J Q K of four suits S H D C.

In draw poker, 5 cards are dealt to each player. Hands are valued in increasing degree as follows.

**Edit:**Examples of each type of hand shown in red.

0 - High card: Highest ranking card. This hand is none of the following types. - 3D 6H 8C JH KD

1 - Pair: 2 cards of same rank; 3 cards of different ranks - 4C 4D x x x

2 - Two Pair: 2 cards of one rank, 2 cards of another rank, 1 card of different rank. - 6H 6S 9D 9H x

3 - Three of a kind: 3 cards of same rank; 2 cards of different ranks. - KC KD KS x x

4 - Straight: 5 cards in rank sequence, but not all of same suit. - 3D 4D 5C 6H 7S

5 - Flush: 5 cards all of the same suit, but not all in sequence. - 3H 6H 7H 9H JH

6 - Full House: 3 cards of one rank; 2 cards of another rank. - AH AC AS 8S 8H

7 - Four of a kind: 4 cards of same rank. 7S 7H 7D 7C x

8 - Straight flush: 5 cards in sequence and all of same suit. - 4H 5H 6H 7H 8H.

Prove that the valuation of 0-8 is reasonable by computing the probabilities of being dealt each

**type**of hand.

Express the probabilities as fractions where the denominator is the number of possible unique 5-card hands dealt from a single deck.

Easy check - they add up to unity.

The prize for being first with all correct answers is 10 "attaboy"s.

The current value of an "attaboy" is that you may combine 100 of them with $0.85 for a cup of coffee at Dunkin Donuts.